Sample Size Determination given CI for proportion. 
Using formula:
$$ n=c^2 \hat p(1-\hat p)/\epsilon^2 $$ where c is 1.96 in this case
I got an answer of 245.8624 for (a), round up to 246.
and 1536.64 for (b), therefore round up to 1537.
I am confused as this hasn't really taken the 10% away from target of 0.8 into account. Should I use 0.7 as my p estimate instead?
Thanks in advance for any help!
Some other thoughts that I have had was maybe to show the info they have given us as p=0.75 with a 5% on either side as a 95% CI??
Think I may have found the answer.
Given that a proportion=0.7 is possible, it is best to use p=0.7 in equation it would cover the possibility of p=0.8, but using p=0.8 would not cover the possibility of p=0.7.
 A: An applied approach to this problem using a Bayesian analysis
I believe your approach and my previous answer are exactly what was
expected by the person who wrote the question. However, viewed as a practical problem about estimating the percentage of "good" bicycles, I find that answer
to be almost entirely unsatisfactory.
We begin this investigation with important prior information that we are invited to use use for planning how many bicycles should be sampled. But even without looking at any bicycles, that information is of considerable importance.
Success probability as a random variable. To a Bayesian statistician, the advance knowledge that the success rate is "never" outside
the interval 80% $\pm$ 10% is crucial prior information to be used directly in investigating the current success rate. In Bayesian statistics the binomial success rate $p$ is modeled as a random variable. Beta distributions are natural distributions to use for $p$ because they have support $(0,1).$ 
Beta prior distribution. In this case, the prior
distribution of this random variable might be something like
$p \sim \mathsf{Beta}(48, 12).$  This particular beta distribution has mean $0.8,$ median $0.803,$ and mode $8.10,$ and it puts about 95% of its probability in the interval $(0.7, 0.8).$ Furthermore, it already yields the
67% probability interval estimate $(0.75, 0.85)$--without doing any 
additional sampling. [Computations in R.]
qbeta(.5, 48, 12)
[1] 0.8033475
diff(pbeta(c(.7, .9), 48, 12))
[1] 0.9522617
diff(pbeta(c(.75, .85), 48, 12))
[1] 0.6699167

One can always quibble about the exact prior distribution to use for a Bayesian analysis, but this one seems
about right. Roughly speaking, if we had begun without any idea of the distribution of $p,$ the facts in the previous paragraph might have resulted
from an inspection of about $60$ bicycles of which about $48$ successfully passed inspection.  
It is difficult to say how many additional bicycles need to be inspected
in order to get an interval estimate that is of length $0.10$ (such as
$(.7, .8), (.75, .85), (.8, .9),$ and so on. That would depend on whether
data from the additional inspection substantially agrees with $p \approx 0.8$ or not. But we can do some intelligent guessing, made somewhat easier by
the track record that $p$ has 'always' been in $(0.7, 0.8).$
Binomial likelihood. One possible scenario: Suppose we tried inspecting $200$ bicycles and that
roughly 80% of them (say 158) pass inspection. Then we have a binomial
likelihood function of the form $p^{158}(1-p)^{42}$ (times an an irrelevant constant). [Likelihood functions are often defined only 'up to a constant multiple'; the important this the functional form involving $p.]$
Beta posterior distribution. Then according to the continuous version of Bayes Theorem, the 'posterior'
distribution of $p$ is the product of the prior density and the likelihood.
written as $$f(p|x) \propto f(p) \times f(x|p),$$
where the symbol $\propto$ [read as "proportional to"] recognizes that we are omitting 'norming constants'. Specifically,
in our case this becomes
$$f(p|x) \propto p^{48-1}(1-p)^{12-1} \times p^{158}(1-p)^{42}
\propto p^{206-1}(1-p)^{54-1},$$
We recognize the result as proportional to the density function
of $\mathsf{Beta}(206, 54),$ which gives the 95% Bayesian posterior
probability interval $(0.74, 0.84),$ approximately of length $0.10.$
qbeta(c(.025,.975), 206, 54)
[1] 0.7410396 0.8393180

Therefore, we see that using the prior information and results from inspecting
only 200 bicycles, we have obtained a 95% probability interval of the desired
length. With different data than in my speculation, the interval estimate
might have been too long. Then you can decide whether it is good enough anyhow, or whether to look at an additional 25 or 50 bicycles. 
Iterative approach. If we took more data
we could use the 'sequential' approach of using the (slightly unsatisfactory)
posteror distribution from the sample of 200 as the prior distribution of
a new analysis. That prior along with the likelihood from an additional few
bicycles gives a new posterior distribution (hopfully more satisfactory).
Concluding comments. If you have never seen a Bayesian analysis before, you may have some questions. This has been an very skeletal account. Maybe you can find answers by searching this site, our sister statistical site ('Crossvalidated'), or the Internet.
Or you might just let it go for now. But be aware that the "obvious"
answer to your simple statistical exercise on confidence intervals would not be used by all statisticians. Not everyone would use a Bayesian analysis. Various other approaches might be used as well.
A: I guess you used the following computation to get $n =246.$
1.96^2 * .8*.2/.05^2
[1] 245.8624

I think your addendum to use 70% to get $n = 323$ is appropriate.
1.96^2 * .5^2/.05^2
[1] 322.6944

If you had no clue as to the value of $p,$ you would use 50% as the
'worst-case scenario'. 
1.96^2 * .5^2/.05^2
[1] 384.16

